- Calculate the moment of inertia of the transformed cracked
section at the center
Referring to Fig. I9e and assuming tentatively that the neutral axis falls within the flange,
we see nAs = 10(3.16) = 31.6 in^2 (203.88 cm^2 ). The static moment with respect to the neu-
tral axis is Q = l/2(68y^2 ) - 31.6(20.5 -y) = O; y = 3.92 in (99.568 mm). The neutral axis
therefore falls within the flange, as assumed. The moment of inertia with respect to the
neutral axis is I 2 = 0/3)68(3.92)3 + 31.6(20.5 - 3.92)^2 = 10,052 in^4 (41.840 dm^4 ). - Calculate the deflection at midspan
Use the equation
L'2 i 5M 1 M 3 \
A=
-Ji (IT-T)(37)
where 7 = average moment of inertia, in^4 (dm^4 ). Thus, /= ^(9737 + 10,052) = 9895 in^4
(41.186 dm^4 ); E = 14515 x 33/c')°^5 = 57,600(250O)^05 = 2,880,000 lb/in^2 (19,857.6 MPa).
Then A = [22^2 x 17287(2880 x 9895)](5 x 200/48 - 100/8) = 0.244 in (6.198 mm).
Where the deflection under sustained loading is to be evaluated, it is necessary to ap-
ply the factors recorded in the ACI Code.
Design of Compression Members by Ultimate-Strength MethodThe notational system is Pu = ultimate axial compressive load on member, Ib (N); Pb =
ultimate axial compressive load at balanced design, Ib (N); P 0 = allowable ultimate axial
compressive load in absence of bending moment, Ib (N); Mu = ultimate bending moment
in member, lb-in (N-m); Mb = ultimate bending moment at balanced design; d' = distance
from exterior surface to centroidal axis of adjacent row of steel bars, in (mm); t = overall
depth of rectangular section or diameter of circular section, in (mm).
A compression member is said to be spirally reinforced if the longitudinal reinforce-
ment is held in position by spiral hooping and tied if this reinforcement is held by means
of intermittent lateral ties.
The presence of a bending moment in a compression member reduces the ultimate ax-
ial load that the member may carry. In compliance with the ACI Code, it is necessary to
design for a minimum bending moment equal to that caused by an eccentricity of 0.05/
for spirally reinforced members and 0.10/ for tied members. Thus, every compression
member that is designed by the ultimate-strength method must be treated as a beam col-
umn. This type of member is considered to be in balanced design if failure would be char-
acterized by the simultaneous crushing of the concrete, which is assumed to occur when
ec = 0.003, and incipient yielding of the tension steel, which occurs when^ =fy. The ACI
Code set </> = 0.75 for spirally reinforced members and <f) = 0.70 for tied members.
ANALYSIS OFA RECTANGULAR MEMBER
BY INTERACTION DIAGRAM
A short tied member having the cross section shown in Fig. 2Oa is to resist an axial load
and a bending moment that induces compression at A and tension at B. The member is
made of 3000-lb/in^2 (20,685-kPa) concrete, and the steel has a yield point of 40,000 lb/in^2
(275,800 kPa). By starting with c = 8 in (203.2 mm) and assigning progressively higher
values to c, construct the interaction diagram for this member.